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I was wondering if anyone could help me solve this probability question to help me make a decision about something I am working on.

In a pool of 2,560,000,000,000,000,000 possible values, if we take 2 random samples from the pool - one sample with 50MM values and another sample with 100MM values - what is the probability that the 2 samples would contain any of the same values?

Any takers?

Thanks Karen

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  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ Oct 2, 2015 at 18:40
  • $\begingroup$ No this is actually work related. Unfortunately that college statistics class I took was over 20 years ago :) $\endgroup$
    – Karen S.
    Oct 2, 2015 at 18:47

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welcome to the sight.

More generally suppose I have $n$ tennis balls and I paint $m$ of them red. Now I'm going to randomly take $l$ balls and hope that none of them are red.

In order to get no red balls I need the first one not to be red. The chances of that are simple enough (I'm assuming we are sampling uniformly) $$ \frac{n-m}{n}. $$

Now, assuming I was successful on the first draw there is one less regular ball, and one less ball in total. So for the next draw, my chances of drawing another regular one is, $$ \frac{n-m-1}{n-1}. $$

If we continue this we can get the probability of the $i^{th}$ ball being regular given all the previous balls were regular. We can compute the probability that the whole sample is regular by taking the product. Generally speaking the product of the probability of one event, and the probability of another event given the first occurred, is the probability that they both occur. This can be chained. So in the end we have: $$ \frac{(n-m)(n-m-1)\dots (n-m-l+1)}{(n)(n-1)\dots (n-l+1)}. $$

Another way of arriving at this solution is by counting. The binomial coefficient, $\binom{n}{k}$, counts the number of ways of select $k$ items from a set of $n$ items. So there are $\binom{n-m}{l}$ ways of choosing $l$ regular balls, and $\binom{n}{l}$ ways of choosing $l$ balls in general. Since each choice is equally likely the probability of getting regular ones is, $$ \frac{\binom{n-m}{l}}{\binom{n}{l}}. $$

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